Problem: The geometric sequence $(a_i)$ is defined by the formula: $a_1 = -\dfrac{16}{3}$ $a_i = \dfrac{3}{4}a_{i-1}$ What is $a_{3}$, the third term in the sequence?
From the given formula, we can see that the first term of the sequence is $-\dfrac{16}{3}$ and the common ratio is $\dfrac{3}{4}$ To find the third term, we can rewrite the given recurrence as an explicit formula. The general form for a geometric sequence is $a_i = a_1 r^{i - 1}$ . In this case, we have $a_i = -\dfrac{16}{3} \left(\dfrac{3}{4}\right)^{i - 1}$ To find $a_{3}$ , we can simply substitute $i = 3$ into the formula. Therefore, the third term is equal to $a_{3} = -\dfrac{16}{3} \left(\dfrac{3}{4}\right)^{3 - 1} = -3$.